Last 4 terms are trivial. But I have trouble following even the first step. The SE answer is also pretty comprehensible, but as if there was some default assumption I’m
not aware of. Do we make one-time substitution, or recursive? Stopping rules feel arbitrary in all enumeration combinatorics I try.
The first step proceeds as follow. We want the predecessor of [s,t] with s=[0,0] and t=0.
We first compute s' = P(s) = P([0,0]) = 0. Then in [s',t] = [0,0] we must replace all occurrences of 0 with [0,0], which results in [[0,0],[0,0]]. This is a one-time substitution (else it would never end).
Comments
Last 4 terms are trivial. But I have trouble following even the first step. The SE answer is also pretty comprehensible, but as if there was some default assumption I’m not aware of. Do we make one-time substitution, or recursive? Stopping rules feel arbitrary in all enumeration combinatorics I try.
https://codegolf.stackexchange.com/a/219466
The first step proceeds as follow. We want the predecessor of [s,t] with s=[0,0] and t=0.
We first compute s' = P(s) = P([0,0]) = 0. Then in [s',t] = [0,0] we must replace all occurrences of 0 with [0,0], which results in [[0,0],[0,0]]. This is a one-time substitution (else it would never end).
Couldn't you also do the substitution n times to make this ordinal emulate the fast growing hierarchy instead of the middle growing one?
Where would you get the number n from? All we have here are binary trees s and t of arbitrary shape.
Ah, now I see, thanks!
Just look at the expanded js code, I also had troubles with the more informal description. It's a one-time substitution but for all matches.